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Connect a bulb marked "10V. 5W" in series with a resistor and connect it to a 10V power supply. The power consumed by the resistor is 1W. Calculate the actual power of (1) bulb and (2) bulb
The resistance of the bulb is r = u & # 178 / P = 10 & # 178 / 5 = 20 (Ω). After being connected in series with the resistance, if the resistance is r, the loop current is I = 10 / (R + 20). Then the power consumed by the resistance is PR = I & # 178; r = 100r / (R + 20) &# 178; = 1 (W). R & # 178; + 40R + 400-100r = 0, R & # 178; - 60R + 400 = 0, so r
Connect a bulb marked "6V 6W" to the power supply, and the actual power of the bulb is 1.5W? (the resistance of the lamp does not change) (2) what is the supply voltage?
(1) The resistance of the bulb: r = U2, P = (6V) 26W = 6 Ω; (2) when the actual power of the bulb is 1.5W, the voltage of the power supply: u = P, r = 1.5W × 6 Ω = 3V
Now there is a "6V, 3W" light bulb and a 9V power supply. In order to make the small light bulb shine normally, (1) how much resistance must be connected in series? How much electric power does the resistance consume?
P=UI
Normal luminous current of small bulb:
I=P/U=3/6=0.5(A)
Total power: P = UI = 9 * 0.5 = 4.5 (W)
Resistance power: 4.5-3 = 1.5 (W)
I=U/R
R=U/I=(9-6)/0.5=6(Ω)
Physics;; a 6V 3W lamp is connected to 9V power supply. In order to make it shine normally, what is the resistance in series? What is the electric power consumed by this resistance
R = u ^ 2 / P = (6V) ^ 2 / 3W = 12 Ω
R = u ^ 2 / P = (9V) ^ 2 / 3W = 27 Ω
27 Ω - 12 Ω = 15 Ω
I=P/U=3W/6V=0.5A
P = I ^ 2R = (0.5A) ^ 2 * 15 Ω = 3.75w
A: a 15 ohm resistor in series consumes 3.75w of electric power
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